What is an interquartile range?

What is a mean absolute deviation?

http://www.mathsisfun.com/data/quartiles.html

http://www.mathsisfun.com/data/mean-deviation.html

49,62, 75, 43, 46

The interquartile range and mean absolute deviation are two measures of dispersion used in statistics to describe the spread or variability of a dataset.

1. Interquartile Range (IQR):
The interquartile range is a measure of the spread of the middle 50% of a dataset. It is computed as the difference between the upper quartile (Q3) and the lower quartile (Q1). Quartiles divide the dataset into four equal parts, where Q1 represents the boundary below which 25% of the data falls, and Q3 represents the boundary below which 75% of the data falls.

To find the interquartile range:
1. Sort the dataset in ascending order.
2. Find Q1 by calculating the median of the lower half of the dataset.
3. Find Q3 by calculating the median of the upper half of the dataset.
4. Calculate IQR by subtracting Q1 from Q3.

For example, given the dataset [1, 2, 3, 4, 5, 7, 8, 9, 9, 10], we would find:
Q1 = 3
Q3 = 9
IQR = Q3 - Q1 = 9 - 3 = 6

2. Mean Absolute Deviation (MAD):
The mean absolute deviation measures the average amount by which each data point in a dataset deviates from the mean. It is the average absolute difference between each data point and the mean of the dataset.

To find the mean absolute deviation:
1. Calculate the mean (average) of the dataset.
2. Find the absolute difference between each data point and the mean.
3. Calculate the average (mean) of the absolute differences.

For example, let's consider the dataset [1, 2, 3, 4, 5, 7, 8, 9, 9, 10]:
1. Find the mean: (1 + 2 + 3 + 4 + 5 + 7 + 8 + 9 + 9 + 10) / 10 = 5.8
2. Calculate the absolute difference for each data point:
|1 - 5.8| = 4.8
|2 - 5.8| = 3.8
|3 - 5.8| = 2.8
|4 - 5.8| = 1.8
|5 - 5.8| = 0.8
|7 - 5.8| = 1.2
|8 - 5.8| = 2.2
|9 - 5.8| = 3.2
|9 - 5.8| = 3.2
|10 - 5.8| = 4.2
3. Calculate the mean absolute deviation: (4.8 + 3.8 + 2.8 + 1.8 + 0.8 + 1.2 + 2.2 + 3.2 + 3.2 + 4.2) / 10 = 2.9

So, the mean absolute deviation for this dataset is 2.9.